# reordering the indices of a matrix

Let $A$ be an $n \times n$ matrix of rank r. Then by reordering the indices if necessary we can bring the matrix in the form $(\frac{A_1}{A_2})$ where $A_1$ is an $r \times n$ matrix, $A_2$ is an $n-r \times n$ matrix and $A_1$ has rank r.

similarly by reordering the indices we can write the matrix $A$ as a direct sum of in decomposable matrices.

I am just thinking how to prove these two facts.

{{{{{ Note: The term irreducible is usually used instead of indecomposable.

Wikipedia: "...a matrix is irreducible if it is not similar via a permutation to a block upper triangular matrix (that has more than one block of positive size)." (Replacing non-zero entries in the matrix by one, and viewing the matrix as the adjacency matrix of a directed graph, the matrix is irreducible if and only if the digraph is irreducible.)

PlanetMath: reducible matrix "An n×n matrix A is said to be a reducible matrix if and only if for some permutation matrix P, the matrix PTAP is block upper triangular matrix." If a square matrix is not reducible, it is said to be an irreducible matrix }}}}}

When we reordering the indices actually what we are doing geometrically or algebraically about the matrix?

Thanks for your valuable time and interest.

You just need to choose $r$ linearly independent rows and move them to the top. That would be left-multiplying by an $n \times n$ permutation matrix.
• nice. similarly how to bring $A$ into block diagonal form by reordering the indices? – GA316 May 27 '15 at 3:00
• @GA316 What do you mean by block diagonal form? Do you mean you want an invertible $r \times r$ submatrix in the top left? If so, choose $r$ columns such that those columns of $A_1$ are linearly independent, and move them to the left. Do so by multiplying on the right by a permutation matrix. – Eric Auld May 27 '15 at 3:05